Problem: Determine how many solutions exist for the system of equations. ${2x-2y = 8}$ ${-x+y = -3}$
Solution: Convert both equations to slope-intercept form: ${2x-2y = 8}$ $2x{-2x} - 2y = 8{-2x}$ $-2y = 8-2x$ $y = -4+x$ ${y = x-4}$ ${-x+y = -3}$ $-x{+x} + y = -3{+x}$ $y = -3+x$ ${y = x-3}$ Just by looking at both equations in slope-intercept form, what can you determine? ${y = x-4}$ ${y = x-3}$ Both equations have the same slope with different y-intercepts. This means the equations are parallel. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ Parallel lines never intersect, thus there are NO SOLUTIONS.